int frac{(x-1) e^{x}}{(x+1)^{3}} d x is equal | vedclass

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(B) Let $I = \int \frac{x-1}{(x+1)^{3}} e^{x} d x$. We can rewrite the numerator as $(x+1)-2$: $I = \int \frac{(x+1)-2}{(x+1)^{3}} e^{x} d x$ $I = \in

Vedclass · Accepted Answer

$\frac{e^{x}}{(x+1)^{2}}+C$

Vedclass · Answer

(B) Let $I = \int \frac{x-1}{(x+1)^{3}} e^{x} d x$. We can rewrite the numerator as $(x+1)-2$: $I = \int \frac{(x+1)-2}{(x+1)^{3}} e^{x} d x$ $I = \int \left( \frac{x+1}{(x+1)^{3}} - \frac{2}{(x+1)^{3}} \right) e^{x} d x$ $I = \int \frac{e^{x}}{(x+1)^{2}} d x - \int \frac{2 e^{x}}{(x+1)^{3}} d x$. Now,apply integration by parts to the first integral $\int \frac{e^{x}}{(x+1)^{2}} d x$,taking $u = \frac{1}{(x+1)^{2}}$ and $dv = e^{x} dx$: $du = -2(x+1)^{-3} dx$ and $v = e^{x}$. $\int \frac{e^{x}}{(x+1)^{2}} d x = \frac{e^{x}}{(x+1)^{2}} - \int e^{x} \left( -\frac{2}{(x+1)^{3}} \right) d x$ $= \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x$. Substituting this back into the expression for $I$: $I = \left( \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x \right) - \int \frac{2 e^{x}}{(x+1)^{3}} d x$ $I = \frac{e^{x}}{(x+1)^{2}} + C$.

$\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x$ is equal to

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